0
$\begingroup$

I have the following statements

At first, theorem 1:
Let
$A$ be a complete metric space,
$B$ a separable metric space,
$M$ in $A\times B$ open and dense set, then

the room $A$ contains a dense set $N$ with the property, that for every point $a$ of $N$ in $B$ exists a dense set of points $y$, such that all points $(a,y)$ belongs to $M$. End theorem 1.

Let $F_n(X):=\{f:X\rightarrow \mathbb{R}^n \;|\;f\;continuous \}$ the space of continuous functions with mapping to $\mathbb{R}^n$ on a compact metric space $X$ provided with sup-metric and
$G_n(X):=\{f:X\rightarrow \mathbb{R}^n \;|\; f\; continuous,\; f(r)\neq 0, \; \forall \; x\in X\}$ the space of continuous functions with mapping to $\mathbb{R}^n$ without functions with zeros, on a compact metric space $X$ provided with sup-metric.

So $F_n(X)$ is a complete, separable metric space, $G_n(X)$ is open in $F_n(X)$ and we can write $F_n(X)$ as $F_1(X)\times F_{n-1}(X)$.

So if $G_n(X)$ is dense in $F_n(X)$ we can use theorem 1.
Therefore there exists a everywhere dense of functions $f\in F_1(X)$ such that:
a) For all functions $\varphi\in F_{n-1}$ who combined with $f$ yield an element $(f,\varphi)\in G_n(X)$, form a dense set in $F_{n-1}$

If we define for $f\in F_1(X)$ with $E(f)$ as set of points $x\in X$ with $f(x)=0$, then for $\varphi\in F_{n-1}$ is the relation $(f,\varphi)\in G_n(X)$ equivalent, that no point of $E(f)$ will be mapped through $\varphi$ in the point $(0,0,\ldots, 0)\in \mathbb{R}^{n-1}$.

I don't understand why $\varphi(x)$ with $x\in E(f)$ can't be mapped to the zero vector. Because, let us start with $F_1(X)\times F_1(X)$. Have a look a my picture (sorry for ugly writing). Visualisation of my imagination

so if we take one constant $f$ in $G_2(X)$ e.g. $f(x)=1$, then $\varphi$ can be mapped to $0$ in $G_2$ (because the constraint that functions with zeros in $G_2$ are not allowed are not broken) and therefore it is allowed to use $\varphi(x)$ with $x\in E(f)$ if $f$ is not a function with zero.

Can anyone help me to find my logic mistake? Thanks

$\endgroup$
1
$\begingroup$

Therefore there exists a everywhere dense of functions $f\in F_1(X)$ such that:
a) For all functions $\varphi\in F_{n-1}$ who combined with $f$ yield an element $(f,\varphi)\in G_n(X)$, form a dense set in $F_{n-1}$

That is there exists a dense set $N$ of $F_1(x)$ such that for each $f\in N$ the set $G_f=\{\varphi\in F_{n-1}(X):(f,\varphi)\in G_{n}(X)\}$ is dense. Note that the set $G_f$ depends on the function $f$ and it may be not the same for distinct $f\in F_1(X)$.

If we define for $f\in F_1(X)$ with $E(f)$ as set of points $x\in X$ with $f(x)=0$, then for $\varphi\in F_{n-1}$ is the relation $(f,\varphi)\in G_n(X)$ equivalent, that no point of $E(f)$ will be mapped through $\varphi$ in the point $(0,0,\ldots, 0)\in \mathbb{R}^{n-1}$.

Right.

I don't understand why $\varphi(x)$ with $x\in E(f)$ can't be mapped to the zero vector.

Note, that we still speak about pairs $(f,\varphi)\in G_n(X)$. And if $x\in E(f)$ and $\varphi(x)=0$ then $(f(x),\varphi(x))=0$, which contradicts $(f,\varphi)\in G_n(X)$.

so if we take one constant $f$ in $G_2(X)$ e.g. $f(x)=1$, then $\varphi$ can be mapped to $0$ in $G_2$ (because the constraint that functions with zeros in $G_2$ are not allowed are not broken) and therefore it is allowed to use $\varphi(x)$ with $x\in E(f)$ if $f$ is not a function with zero.

We are allowed to use $\varphi$ means $(f,\varphi)\in G_n(X)$ for a given fixed $f$. This doesn’t imply that we are allowed to use $\varphi$ with a different function, say $f’\in F_1(X)$. For a particular choice of $f$, when $f(x)=1$ for each $x\in X$, a set $E(f)=\{x\in X :f(x)=0\}$ is empty. Then $G_f=\{\varphi\in F_{n-1}(X):(f,\varphi)\in G_{n}(X)\}=F_{n-1}(X)$. But this does not mean that $G_{f’}=F_{n-1}(X)$ for any function $f’\in F_1(X).$

Remark that in this case also $E(f)=\{x\in X :f(x)=0\}$ is empty, so we cannot "use $\varphi(x)$ with $x\in E(f)$", because there is no such $x$. But if $f’\in F_1(X)$ is an other function then we are allowed to use a function $\varphi\in F_{n-1}$ with the function $f’$, that is $(f,\varphi)\in G_n(X)$ iff $\varphi(x)\ne 0$ for each $x\in E(f’)$.

PS. It seems that the picture is not needed.

$\endgroup$
9
  • $\begingroup$ And what about if only one $f$ is constant e.g. $f=1$ and the other have zeros. So one can write $(1,\varphi)$ with $\varphi$ abitrary. Then it did not contradicts $(f,\varphi)\in G_n(X)$? $\endgroup$ – GeoRie Dec 6 '18 at 14:28
  • $\begingroup$ @GeoRie Looking for $(f,\varphi)\in G_n(X)$, we cannot pair arbitrary $f$ with arbitrary $\varphi$, we need a restriction $\varphi(x)\ne(0,0,\dots,0)\in\Bbb R^{n-1}$ for each $x\in E(f)$, right? $\endgroup$ – Alex Ravsky Dec 6 '18 at 16:09
  • $\begingroup$ Yes. That's my question. Why we need a restriction to $\varphi$ if we combine it with those $f$ who did not have zeros. Then we don't need a restriction for $\varphi$? Thrn, if we choose such $f$, $\varphi$ can be mapped to zero in $\mathbb{R}$? $\endgroup$ – GeoRie Dec 7 '18 at 9:21
  • $\begingroup$ Do you understand my thoughts? Are my thoughts wrong? Could you explain it more exact, please. Thanks a lot. $\endgroup$ – GeoRie Dec 9 '18 at 11:40
  • $\begingroup$ @GeoRie I expanded my answer. $\endgroup$ – Alex Ravsky Dec 9 '18 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.